Question

A ball is released at point P with a tangential velocity of 5 ms-1
to move in a circular track in a
vertical plane as shown in the Figure 1. Can the ball reach the highest point of the circular track
of radius 1.0 m?

Answer 1

Assume that there is no friction.

$\frac{mv^2}{2}=mgh\to h=\frac{v^2}{2g}=\frac{5^2}{2\cdot9.81}=1.27(m)$

In our case $h=2R=2\cdot1=2(m)$.

$2>1.27$

So, the ball can not reach the highest point of the circular track.